The answer to the first question is probably disappointingly simple: Take $\mathbb{R}^2$, once with the Euclidean norm $\|(x,y)\|_2 = (|x|^2+|y|^2)^{1/2}$ and once with the $\ell^1$-norm $\|(x,y)\|_1 = |x| + |y|$ for example. The norms are equivalent, the Euclidean norm is uniformly convex, while the $\ell^1$-norm isn't. Of course, $\mathbb{R}^2$ is reflexive with respect to both norms.
The answer your updated second question is a classical result of M.M. Day:
Theorem. There exist Banach spaces which are separable, reflexive, and strictly convex, but are not isomorphic to any uniformly convex space.
This is the main result of his paper bearing the statement of the theorem in its title:
M.M. Day, Reflexive Banach spaces not isomorphic to uniformly convex spaces, Bull. Amer. Math. Soc. Volume 47, Number 4 (1941), 313-317, MR0003446.
Since the paper is freely available via the above link, there is little sense in my elaborating on it.
Larsen's Functional Analysis, an Introduction, Thm 8.2.2 has an elementary proof of this, that relies on a lemma, which he leaves to the reader. Here is the statement and proof of the lemma:
Let $E$ be a uniformly convex normed linear space over $F$. If $(x_n)\subseteq E$ is a sequence such that
$(i) \underset{n\to \infty}\lim \|x_n\| = 1,\ (ii) \ \underset{n,m\to \infty}\lim \|x_n + x_m\| = 2.\ \text{Then},\ (x_n)\ \text{is Cauchy.}$
$(i)$ implies that for each integer $k$, there is an $x_{n_k}$ such that $\|x_{n_k}\|<1+1/k$.
Now, if $(x_n)$ is not Cauchy, then we may choose $(x_{n_k})$ so that the subsequence is not Cauchy either. Then, there is an $\epsilon>0$ such that for all integers $K$, there are integers $n_k,n_l$ such that $\|x_{n_k}-x_{n_l}\|>\epsilon$ with $k,l>K.$
Without loss of generality, assume $l>k$ so that $1+1/l<1+1/k$ and therefore $\|x_{n_k}\|<1+1/k$ and $\|x_{n_l}\|<1+1/l<1+1/k.$
Then, the hypothesis of uniform convexity gives us a $\delta>0$ such that $\frac{\|x_{n_k}+x_{n_l}\|}{2(1+1/k)}<1-\delta.$ But now, on taking limits, and applying $(ii)$,we get a contradiction: $1=\underset{k,l\to \infty}\lim\frac{\|x_{n_k}+x_{n_l}\|}{2(1+1/k)}\le 1-\delta$
Best Answer
If $x$ is not a multiple of $y$, $\Vert y\Vert\ge\Vert x\Vert>0$, and $\Vert x+y\Vert=\Vert x\Vert+\Vert y\Vert$, then $$ \biggl\Vert {x\over\Vert x\Vert}+ {y\over\Vert y\Vert} \biggr\Vert \ge \biggl\Vert {x\over\Vert x\Vert}+ {y\over\Vert x\Vert} \biggr\Vert - \biggl\Vert {y\over\Vert x\Vert}- {y\over\Vert y\Vert} \biggr\Vert ={{\Vert x\Vert +\Vert y\Vert}\over \Vert x\Vert} -\Vert y\Vert\Bigl({1\over\Vert x\Vert}-{1\over \Vert y\Vert}\Bigr)=2. $$
So $X$ is strictly convex if and only if for distinct norm one vectors $x$, $y$, $\Vert x+y\Vert<2$.
$X$ is uniformly convex if and only if for each $\epsilon>0$, there is a $\delta>0$ so that for any $x$, $y$ in the closed unit ball of $X$ with $\Vert x-y\Vert\ge\epsilon$ one has $$\Bigl\Vert{ x+y\over 2}\Bigr\Vert\le 1-\delta.$$
It should be easy to see from the above that a uniformly convex normed space is strictly convex.